# What does “standard Koszul morphism” mean?

I'm reading a paper 'D'Andrea, Carlos(RA-UBA), Dickenstein, Alicia(RA-UBA)Explicit formulas for the multivariate resultant. (English summary) Effective methods in algebraic geometry (Bath, 2000). J. Pure Appl. Algebra 164 (2001), no. 1-2, 59–86.'

Let $$f_1,\ldots,f_n$$ be homogeneous polynomials of degrees $$d_1,\ldots,d_n$$ and $$A=\mathbb Z[a_{\alpha}]$$, where $$a_{\alpha}$$ denotes the all coefficients of $$f_1,\ldots,f_n$$. Let $$S_m$$ be a $$A$$-free module generated by the monomials in $$A[x_1,\ldots,x_n]$$ with degree $$m$$.

In p.73, the author define a Koszul complex $$K^{\bullet}(t;f_1,\ldots,f_n)$$ by

$$\{ 0 \to K(t)^{-n} \to \cdots \to K(t)^{-1} \to K(t)^0 \},$$

where

$$K(t)^{-j}=\bigoplus_{i_1 < \cdots < i_j} S_{t-d_{i_1}-\cdots-d_{i_j}},$$

and the morphisms from $$K(t)^{-j}$$ to $$K(t)^{-j+1}$$ are the standard Koszul morphisms.

Here what does "standard Koszul morphism" mean? I couldn't catch it.

• To $f_i$ corresponds an element, say $F_i$, of $S_{d_i}$. Multiplying by $F_i$ one gets complexes $0\to S_m \to S_{m+d_i}\to 0$. Now try a tensor product of such complexes. – Wilberd van der Kallen Aug 8 at 7:21
• Wilberd van der Kallen // Could you let me know details of 'try a tensor product of such complexes?' – LWW Aug 8 at 7:31
• en.wikipedia.org/wiki/Koszul_complex – Denis Nardin Aug 8 at 8:21
• Nardin // My question is that, tensor product means tensor product over i=1,...,n? – LWW Aug 8 at 9:25